Claim. For $p$ a prime, let $\Sigma\le S_{p-1}$ be a abelian subgroup of order $p-1$, such that, for every $\sigma\in\Sigma$ and $i\in\{1,\dots,p-1\}$: $$\sigma(i)\equiv\sigma(1)i\pmod p \tag1$$ Then $\Sigma$ is cyclic. $\space\Box$
For example, for $p=5$, the only $\Sigma$ complying with $(1)$ is $\Sigma=\langle (1243)\rangle$. For $p=7$ and $p=11$ there's nothing to prove, provided that one knows that the only abelian groups of order $6$ and $10$ are cyclic. Going up, the first nontrivial (to me) cases are then $p=13$ and $p=17$.
Is there any permutational proof of this claim? I've been trying to contradict $(1)$ by assuming that there is a subgroup $H\le \Sigma$ isomorphic to $C_q\times C_q$, where $q$ is a prime. I can do that, but not by bare permutation arguments.
You can do this with any group. Let $G$ be a finite group and $\Pi(G)$ the group of permutations of elements of $G$. Then you have an injective homomorphism $G\mapsto \Pi(G)$ induced by left multiplication. $$\sigma_g(h) = gh$$ and since $\sigma_g(1) = g$ you can write this as $$\sigma_g(h) = \sigma_g(1) h$$ The identity $\sigma_g(1)=g$ also shows that the image of $G$ acts transitively.